Problem 13 - Olympiad

The system has no solution when the coefficient matrix is singular (determinant = 0) but the augmented matrix has a different rank. The coefficient matrix is A = [[2,3,1],[1,k,1],[3,5,k]]. Compute |A|: |A| = 2(k·k - 1·5) - 3(1·k - 1·3) + 1(1·5 - k·3) = 2(k^2 - 5) - 3(k - 3) + (5 - 3k) = 2k^2 - 10 - 3k + 9 + 5 - 3k = 2k^2 - 6k + 4. Setting |A| = 0: 2k^2 - 6k + 4 = 0 => k^2 - 3k + 2 = 0 => (k-1)(k-2) = 0. So k = 1 or k = 2. We need to check which value makes the system inconsistent. For k = 1: A = [[2,3,1],[1,1,1],[3,5,1]]. Row reduce: R2 = R2 - (1/2)R1, R3 = R3 - (3/2)R1. This gives [[2,3,1],[0,-1/2,1/2],[0,-2/2, -1/2]]. R3 = R3 - 2·R2: [[2,3,1],[0,-1/2,1/2],[0,0, -3/2]]. The coefficient matrix has rank 3, so |A| ≠ 0 for k=1. Wait, but we found |A|=0 for k=1. Let me recompute |A| for k=1: |A| = 2(1-5) - 3(1-3) + (5-3) = 2(-4) - 3(-2) + 2 = -8 + 6 + 2 = 0. Yes. But row reduction suggests rank 3. Let me redo the row reduction carefully for k=1. A = [[2,3,1],[1,1,1],[3,5,1]]. R2 = R2 - (1/2)R1: [1-1, 1-1.5, 1-0.5] = [0, -0.5, 0.5]. R3 = R3 - (3/2)R1: [3-3, 5-4.5, 1-1.5] = [0, 0.5, -0.5]. So we have [[2,3,1],[0,-0.5,0.5],[0,0.5,-0.5]]. Now R3 = R3 + R2: [0, 0, 0]. So rank is 2. Now check the augmented matrix for k=1: Augmented = [[2,3,1,1],[1,1,1,0],[3,5,1,2]]. After same row operations: R2 becomes [0,-0.5,0.5, -0.5] (since 0 - 0.5·1 = -0.5). R3 becomes [0,0.5,-0.5, 0.5] (since 2 - 1.5·1 = 0.5). Then R3 = R3 + R2: [0,0,0,0]. The last row of the augmented matrix is [0,0,0,0], so the system is consistent (infinitely many solutions). For k = 2: A = [[2,3,1],[1,2,1],[3,5,2]]. |A| = 2(4-5) - 3(2-3) + (5-6) = 2(-1) - 3(-1) + (-1) = -2 + 3 - 1 = 0. Row reduce: R2 = R2 - (1/2)R1: [0, 0.5, 0.5]. R3 = R3 - (3/2)R1: [0, -0.5, 0.5]. R3 = R3 + R2: [0,0,1]. So rank of A is 3? But |A|=0, so rank should be < 3. Let me recompute: R2: [1-1, 2-1.5, 1-0.5] = [0, 0.5, 0.5]. R3: [3-3, 5-4.5, 2-1.5] = [0, 0.5, 0.5]. So R3 - R2 = [0,0,0]. So rank is 2. Augmented matrix for k=2: [[2,3,1,1],[1,2,1,0],[3,5,2,2]]. R2 becomes [0,0.5,0.5, -0.5]. R3 becomes [0,0.5,0.5, 0.5]. R3 - R2 = [0,0,0,1]. This gives [0,0,0,1], which is inconsistent. Therefore, for k = 2, the system has no solution.